Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 6x}{x + 6} = \dfrac{-2x - 12}{x + 6}$
Explanation: Multiply both sides by $x + 6$ $ \dfrac{x^2 + 6x}{x + 6} (x + 6) = \dfrac{-2x - 12}{x + 6} (x + 6)$ $ x^2 + 6x = -2x - 12$ Subtract $-2x - 12$ from both sides: $ x^2 + 6x - (-2x - 12) = -2x - 12 - (-2x - 12)$ $ x^2 + 6x + 2x + 12 = 0$ $ x^2 + 8x + 12 = 0$ Factor the expression: $ (x + 6)(x + 2) = 0$ Therefore $x = -6$ or $x = -2$ At $x = -6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -6$, it is an extraneous solution.